B Conservative forces and conservation of mechanical energy

Why do conservative forces conserve mechanical energy while non conservative forces do not?
According to me,
What makes the conservative forces path independent is that for a particular case they always act in a fixed direction irrespective of the direction of motion of the object on which they are acting while non conservative forces change their direction depending on the direction of motion of the object on which they are acting.

Thus in case of conservative forces if an object first moves in the opposite direction of the conservative force "some positive work" is done and then if the object moves in the direction of the conservative force the "some positive work" done before gets neutralized by "some negative work" and for perpendicular motion of the object zero work is done. Thus the potential energy at a point is the same irrespective of the path taken since ultimately the work done is accounted for the change initial and final positions of the object only. Gravitational force is an example of conservative force.

What makes the non-conservative forces path dependent is that for a particular case they always change their direction according to the motion of the object on which they are acting such that the relative orientation of the motion of the object and the non-conservative force remains the same.Thus the work done by the non-conservative forces has the same sign throughout and adds up to give different values of work done for different paths taken unlike conservative force for which the work done which may have different signs in the course of motion which cancel(neutralize each other out). Frictional force is an example of non-conservative force.
Am i correct here?Are all conservative forces unidirectional while non-conservative forces are multidirectional,i.e., is directionality a measure of whether the forces are conservative or non-conservative?

No. One of the most important potentials is the Kepler potential,
$$V(\vec{x})=-\frac{G m M}{r}, $$
where ##G## is the Newton's gravitational constant, ##m## and ##M## the mass of the planet and the sun (which is assumed to be at rest in the origin of the coordinate system), and ##r=|\vec{x}|##. The force is
$$\vec{F}=-\vec{\nabla} V=-\frac{G m M}{r^3} \vec{x}.$$
This is not unidirectional but radially symmetric around the origin.

Staff: Mentor

@Dale, a conservative force conserves mechanical energy and for the same a corresponding potential can be defined. Agreed. But why does a conservative force conserve mechanical energy while a non-conservative force does not. What makes them do so?

@Dale, a conservative force conserves mechanical energy and for the same a corresponding potential can be defined. Agreed. But why does a conservative force conserve mechanical energy while a non-conservative force does not. What makes them do so?

If an object traverses a particular path while acted upon by a conservative force, the work done by that force over that path is the same as the work done by that force over any alternate path. Agreed?

Pithier answer: For the same reason that 2+2=4: It follows from the definitions of the terms.

Well i think that i should rephrase the question: My actual question is why do some forces conserve mechanical energy while others do not?

Well it kind of is in the definition of work. Some forces don't really care about the path you take, you always end up with the same energy as long as the end points are the same [correction from: the path is the same]. Mathematically, in vector calculus it is a consequence of stoke's theorem.
Physically, some forces convert kinetic energy into heat, sound and transformation, and usually the process is irreversible, thus energy not conserved.

Staff: Mentor

But why does a conservative force conserve mechanical energy while a non-conservative force does not. What makes them do so?

I don't understand your repetition of the question. You have already asked this same question multiple times and received the same answer multiple times. What more can be done?

Not all forces conserve energy. Those that do are called conservative and those that do not are not called conservative. Asking what makes a conservative force conserve energy is like asking why an isolated system has no external force. It is a definition.

You proposed that you can identify conservative forces by the fact that they are unidirectional. @vanhees71 showed that is not correct, by counterexample. I instead explained that you can identify conservative forces by the fact that they have a potential.

Conservative forces conserve energy and have a potential. Not all forces conserve energy, not all forces have a potential. Forces that have a potential conserve energy and are called conservative.

Please don't repeat the same question again. It is highly frustrating. If the provided responses don't answer what you want then you need to ask a different question.

@Dale, I had reworded the question in one of replies above and by the way the question i had asked earlier was How do conservative forces conserve mechanical energy while my reworded question reads: Why do some forces conserve mechanical energy while others do not?

Although B-tagged, I give the math. Without the math you have no chance to understand what we are talking about anyway. By definition a force is called conservative, if there exists a scalar potential for it, i.e.,
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
which also doesn't depend explicitly on time.

Now for a point particle you have Newton's equation of motion,
$$m \ddot{\vec{x}}=\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}).$$
To derive that in this case total energy is conserved, take the scalar product of this equation with ##\dot{\vec{x}}##. On the left-hand side you get, using the product rule of differential calculus
$$m \ddot{\vec{x}} \cdot \dot{\vec{x}}=\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t} \dot{\vec{x}}^2.$$
On the right-hand side you get with the chain rule of differentiation
$$-\dot{\vec{x}} \cdot \vec{\nabla} V = -\frac{\mathrm{d}}{\mathrm{d} t} V.$$
Thus you finally get
$$\frac{m}{2} \frac{\mathrm{d}}{\mathrm{d} t} \dot{\vec{x}}^2=-\frac{\mathrm{d}}{\mathrm{d} t} V,$$
or bringing the expression with the potential to the left-hand side
$$\frac{\mathrm{d}}{\mathrm{d}t} \left (\frac{m}{2} \dot{\vec{x}}^2 + V \right)=0 \; \Rightarrow \; E=\frac{m}{2} \dot{\vec{x}}^2 + V=\text{const}.$$
So the total energy is conserved in this case, i.e., if the force has a potential that is not explicitly time dependent.